What is the solution for x^2y^{''}+2xy^'-2=0 ?

The solution for x^2y^{''}+2xy^'-2=0 is y=c_{1}+(c_{2})/x+2ln(x)-2

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x^2y^{''}+2xy^'-2=0

x^{2} y^{^{''}} + 2 x y^{^{'}} - 2 = 0

y = c_{1} + \frac{c _{2}}{x} + 2 ln (x) - 2

Solution steps

x^{2} y^{''} (x) + 2 x y^{'} (x) - 2 = 0

Solve linear ODE:

y = c_{1} + \frac{c _{2}}{x} + 2 ln (x) - 2

y = c_{1} + \frac{c _{2}}{x} + 2 ln (x) - 2

\frac{d}{d x} (y^{2} - x^{2})

\frac{\partial}{\partial x} (x^{0.25} y^{0.5})

\frac{d y}{d x} = x - 3

cos (8 t + 2 π)

\int 7 tan^{3} (x) sec^{3} (x) d x

What is the solution for x^2y^{''}+2xy^'-2=0 ?
The solution for x^2y^{''}+2xy^'-2=0 is y=c_{1}+(c_{2})/x+2ln(x)-2